Question: Let $h(x)=\cot(6x-x^2)$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{\sin^2(2x-6)}$ (Choice B) B $\dfrac{2x-6}{\sin^2(6x-x^2)}$ (Choice C) C $\dfrac{6-2x}{\sin^2(6x-x^2)}$ (Choice D) D $-\dfrac{1}{\sin^2(6x-x^2)}$
$h$ is a trigonometric function, but its argument isn't simply $x$. Therefore, it's a composite trigonometric function. In other words, suppose $u(x)=6x-x^2$, then $h(x)=\cot\Bigl(u(x)\Bigr)$. $h'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[\cot\Bigl(u(x)\Bigr)\right]=-\dfrac{u'(x)}{\sin^2\Bigl(u(x)\Bigr)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned}h'(x) \\\\ &=\dfrac{d}{dx}\cot(6x-x^2) \\\\ &=\dfrac{d}{dx}\cot\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=6x-x^2} \\\\ &=-\dfrac{u'(x)}{\sin^2\Bigl(u(x)\Bigr)} \\\\ &=-\dfrac{6-2x}{\sin^2\Bigl(6x-x^2\Bigr)}&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=\dfrac{2x-6}{\sin^2\Bigl(6x-x^2\Bigr)} \end{aligned}$ In conclusion, $h'(x)=\dfrac{2x-6}{\sin^2\Bigl(6x-x^2\Bigr)}$.